Answer
$P(x)=(x-1)(x+5)(x-2)$
Zeros: $\ \ -5,\ 1, \ 2 \ \ \ $
Work Step by Step
$P(x)=x^{3}+2x^{2}-13x+10$
$P(-x)=-x^{3}+2x^{2}+13x+10$
Decscart's rule of signs:
P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros.
P(-x) has 1 sign change $\Rightarrow$ 1 negative zero.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 5,\pm 10$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 5,\pm 10$
Testing with synthetic division,
$\left.\begin{array}{l}
1|\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 &2&-13&10\\\hline
&1&3 &10\\\hline
1&3&-10&|\ \ 0\end{array}$
Remainder th:
$x=1$ is a zero $\Rightarrow (x-1)$ is a factor (Factor th.).
$P(x)=(x-1)(x^{2}+3x-10)$
For the trinomial, find two factors of $-10$ with sum $3$
... ( they are $+5$ and $-2)$
$P(x)=(x-1)(x+5)(x-2)$
Zeros: $-5,\ 1, \ 2 \ \ \ $
